Asked by elssie
if 25cm3 of a 0.1M NA2CO3 solution neutralised a solution containing 2.5g sulphuric[VI] acid, H2SO4, in 250cm3 of solution, calculate the molarity of H2SO4 and the volume of the acid used
Answers
Answered by
DrBob222
moles H2SO4 = g/molar mass = 2.5/98 = 0.0255
M = moles/L = 0.0255/0.25 L = 0.102
Then Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2
mols Na2CO3 = M x L = 0.1 x 0.025 = 0.0025
mols H2SO4 must also = 0.0025
mols H2SO4 = M x L so 0.0025 = 0.0255 x L. Solve for L.
M = moles/L = 0.0255/0.25 L = 0.102
Then Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2
mols Na2CO3 = M x L = 0.1 x 0.025 = 0.0025
mols H2SO4 must also = 0.0025
mols H2SO4 = M x L so 0.0025 = 0.0255 x L. Solve for L.
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